<a name="docA001">　Index begin　Index this file
2009-06-08-19-10 start
This file http://freeman2.com/tute0008.htm
is Freeman's reading notes. Although Freeman
always keep correct view point. But Freeman's
capability is limited, plus no one proofread
this file. Then you can still find wrong view
point. When read, please put question mark as
often as possible. If you suspect any view 
point wrong, please ask a math expert near by.

<a name="docA002">,<a name="textbook">
This file is a note for reading inequality
book written by Professor J. Michael Steele
The Cauchy-Schwarz Master Class  　★★★★★
Below use 'textbook' as abbreviation.
Freeman also read web pages online, and will 
indicate the source URL at discussion point.

<a name="docA003">
Freeman study mechanical engineering. 
Engineering mathematics do not teach 
inequality. Above book is first inequality
book. First time read, it was very hard.
Although high school time learned
  a*a + b*b >= 2*a*b
But this little knowledge do not help.

<a name="docA004">
This file follow textbook chapter section
order, but not continuous, Freeman skip 
those uncertain sections/problems.

This file first function is to learn 
inequality. Second function is to learn
how to use html code to write math 
equations.

<a name="docA005">　Index begin　Index this file
On 2009-01-27-10-08 Freeman accessed
the next page
http://www.sftw.umac.mo/~fstitl/10mmo/inequality.html
save as sftw.umac.mo_text_math_eqn_good.htm
Above page is the main reference for html
math equation.

In order to let reader to build  html math
equation, previous file tute0007.htm page 
end has math symbol and internal code.

This file third function is to display how
to draw curves in web page. The main engine
is XYGraph v2.3 - Technical Figures

Thank you for read Freeman's inequality page.
2009-06-08-19-47 stop

<a name="docA006">
2009-08-23-15-00 start
□　Exercise 1.14 solution

Exercise 1.14 problem (a) is solved by hint
Exercise 1.14 problem (b) is out of LiuHH's
reach. What is "cardinality of a set B⊂Z3" ?
Without clear understand the definition, 
LiuHH is unable to solve problem (b)
LiuHH major in mechanical engineering.
LiuHH is mathematics admirer and outsider.

<a name="docA007">
To solve problems in
The Cauchy-Schwarz Master Class
LiuHH's goal is to solve 50% of the exercise
problems. In reality, solve 40% is good enough.
Please do not be surprise that LiuHH skip some
text section, skip some problems.
2009-08-23-15-12 stop



<a name="ch06f001">　Index begin　Index this file
2010-01-16-10-46 start
■　Exercise 6.12 problem statement
　　textbook page 103
（The Gauss-Lucas Theorem）
Show that for any complex polynomial
  P(z)=a0+a1z+...+anzn ---eqn.AU001
//Polynomial roots, coefficients.
The roots of the derivative P'(z) are
contained in the convex hull H of the
roots of P(z).
2010-01-16-10-51 stop

<a name="ch06f002">　Index begin　Index this file
2010-01-16-15-23 start
■　Exercise 6.12 hint
　　textbook page 251

If we write
  P(z)=an(z-r1)m1(z-r2)m2...(z-rk)mk ---eqn.AU002
//●●change　textbook use z－rn, 
//LiuHH change to z－rk
<a name="ch06f003">
where r1,r2,...,rk are the distinct 
roots of P(z) and m1,m2,...,mk are 
the corresponding multiplication,
then comparison of P'(z) and P(z)
gives us the familiar formula

<a name="ch06f005">
2010-01-16-15-49 here
Now if z0 is a root of P'(z) which
is also a root of P(z), then z0 is
automatically in H, so without loss of
generality, we may assume that z0 is a
root of P'(z) that is not a root of 
P(z), in which case we find
//LiuHH: in which case z0-rk≠0

<a name="ch06f008">
2010-01-16-16-11 here
textbook use over-bar for complex conjugate.
But text editor can not create over-bar. In 
this file, use superj as complex conjugate 
of super.

<a name="ch06f009">
If we set
  wk=mk/|z0－rk|2 ---eqn.AU006
then we can rewrite this identity as

<a name="ch06f011">
2010-01-16-17-15 here
which shows z0 is a convex combination
of the roots of P(z).
2010-01-16-17-16 stop






<a name="ch06f012">　Index begin　Index this file
2010-01-16-18-01 start
■　Exercise 6.12 solution


Exercise 6.12 hint already solved.
Here make few notes.

If P(z) has repeated roots, these 
roots show up in P'(z).
eqn.AU003 is the ratio of P'(z)/P(z)
z is variable. If z do not equal P(z)
root, then no divide by zero trouble.

<a name="ch06f013">
In eqn.AU004, we set z=z0 and require 
z0 be root of P'(z). We know P'(z) and 
P(z) share roots, if root is repeated.
Whether there is P(z0)=0 risk ? If
P(z0)=0, then eqn.AU004 has divide
by zero trouble.

<a name="ch06f014">
The answer is 'safe'. P(z0)=0 only if
root is repeated. But for repeated 
root it is already in P(z) hull.
No need worry it is in hull or not.
Exercise 6.12 hint excluded repeated
root. Other than repeated root, P'(z)
root and P(z) root are different.
<a name="ch06f015">
Please goto drawing board and click
[12] button.
Figure suggest that for non-repeated
root, P'(z) root and P(z) root are 
different. Division of 
  [P'(z) zero - P(z) zero)]
is safe. Here
 P'(z) zero is z0
 P(z)  zero is r1, r2 etc.

<a name="ch06f016">
eqn.AU004 all denominator/numerator
multiply by denominator complex 
conjugate. All denominator become
real number. Set the following short
representation
  wk=mk/|z0－rk|2 ---eqn.AU006
Here wk is real number, because
root multiplicity mk is real and 
complex * its own conjugate =
|z0－rk|2 is also real.

<a name="ch06f017">　Index begin　Index this file
eqn.AU005 equal to zero (a real number).
We take zero's complex conjugate still
get zero. However the complex number 
in eqn.AU005 will reverse one more time.
From complex conjugate back to original
complex number.

<a name="ch06f018">
In between eqn.AU005 and eqn.AU007 is
next equation 
(with the help of eqn.AU006)
  0 ＝ w1(z0－r1)＋w2(z0－r2)
    ＋...＋wk(z0－rk) ---eqn.AU008
Dropped complex conjugate sign "j" in 
complexj, because reverse one more time.
re-write eqn.AU008 as
    w1z0＋w2z0＋...＋wkz0
  ＝ w1r1＋w2r2＋...＋wkrk ---eqn.AU009

<a name="ch06f019">
From eqn.AU009 to eqn.AU007 is 
simple.

Now we have eqn.AU007 in hand. How
to explain
[[
which shows z0 is a convex combination
of the roots of P(z).
]]
<a name="ch06f020">
eqn.AU007 right hand side are P(z)
roots r1,r2,...,rk (and their weights)
eqn.AU007 left hand side is P'(z)
root z0 [non-repeated in P(z)]

<a name="ch06f021">
Please see 
What is convex set, convex function?
eqn.AU007 tell us that r1,r2,...,rk 
have positive weight wi/∑wk, these
weights sum to one. (i th weight is 
 wi/∑wk. Sum all weights over i get
∑wi/∑wk. i and k are dummy, the 
result is one.) Then z0 is a linear 
combination of r1,r2,...,rk . z0 must
stay within [min(ri), max(ri)]

<a name="ch06f022">
Here showed P'(z) root z0 stay in
P(z) roots hull [min(ri), max(ri)]
min/max is for one-dimension explanation.
2D and 3D case, eqn.AU007 similar to the
gravy center of a body, must stay inside 
of its budy hull. Pinocchio's body center
can not be out side of his nose tip!

(if weights not sum to one, then
 no guarantee)
2010-01-16-19-05 stop



<a name="ch06f023">　Index begin　Index this file
2010-01-16-21-04 start
■　Exercise 6.13 problem statement
　　textbook page 103
（Wilf's Inequality）

Show that if H is the convex hull of
the roots of the complex polynomial
  P(z)=a0+a1z+...+anzn ---eqn.AU010
//Polynomial roots, coefficients.

then one has

<a name="ch06f025">
2010-01-16-21-21 here
where the angle ψ is defined by 
Figure 6.5. This inequality provides
both a new proof and a quantitative 
refinement of the classic Gauss-
Locas Theorem of Exercise 6.12.
2010-01-16-21-24 here






<a name="ch06f026">
2010-01-16-21-26 start
■　Exercise 6.13 hint
　　textbook page 251
Write r1,r2,...,rn for the roots of P
repeated according to their multiplicity,
and for a z which is outside of the
convex hull H write z-rj in polar form
  z-rj = ρj*eiθj ---eqn.AU011
<a name="ch06f027">
we then have
  1/(z-rj) = ρj-1*e-iθj ---eqn.AU012
  1≦j≦n
and the spread in the arguments θj
1≦j≦n is not more than 2ψ. Thus by
the complex AM-GM inequality (2.35)
one has the bound

just as we hoped to prove.
2010-01-16-21-57 stop



<a name="ch06f030">　Index begin　Index this file
2010-01-17-13-51 start
■　Exercise 6.13 solution


Follow hint as next.
Write r1,r2,...,rn for the roots of P
repeated according to their multiplicity.

//2010-01-21-17-42 deleted a numerical
//example section. tute0027.htm is too
//long and numerical example is not in
//textbook. Label ch06f031 to ch06f043
//are missing.

<a name="ch06f044">
Problem require angle
  |ψ| ＜ PI/2 ---eqn.AU014
the reason is stated at Exercise 2.13 
solution. Based on Exercise 2.13
we accept eqn.AU013 is true.

<a name="ch06f045">
From eqn.AU013 to the target equation
eqn.14.54 same as eqn.6.26 
Exercise 6.13 hint give one line
[[
and in terms of P and P', this simply says 
eqn.14.54
]]
How to explain this one line message?
2010-01-17-15-19 here

<a name="ch06f046">　Index begin　Index this file
eqn.AU013 left side numerator cos(ψ) 
move to right side denominator is 
simple but with alert. Because we have
 |ψ| ＜ PI/2 ---eqn.AU014
(Trace back to Exercise 2.13 solution)
cos(ψ)＞0 for ψ in (-PI/2, +PI/2). 
A positive number cos(ψ) move cross
inequality sign NOT reverse the
direction of inequality sign.
Opposite example is here with reason.

<a name="ch06f047">
After moved cos(ψ), next is eqn.AU013
left side absolute term.
How to say that (forget power 1/n)
  1/[(z－r1)*(z－r2)*...*(z－rn)] ---eqn.AU015
is
  an/P(z) ---eqn.AU016
This part is simple.
Polynomial P(z) highest power term 
has coefficient an, see eqn.AU010
<a name="ch06f048">
We factor out an from P(z) 
  P(z)=an*Pb(z) ---eqn.AU017
let new Pb(z) highest power term has 
ONE as its coefficient. Then new Pb(z)
is simply multiplication of all
(z－rk) terms. Since z is variable,
rk is one root. When z step on any 
one rk, (z－rk) become zero. 
<a name="ch06f049">
That is what 'root' for. If not become 
zero, how can we call rk as root ?
eqn.AU015 is 1/Pb(z). From eqn.AU017
  1/Pb(z) = an/P(z) ---eqn.AU018
1/Pb(z) is used in eqn.AU013 //an cancelled
an/P(z) is used in eqn.14.54 //an remain
            and in eqn.6.26
Therefore eqn.AU015 and eqn.AU016 
are same thing.
2010-01-17-16-00 here

<a name="ch06f050">
Next explain eqn.AU013 great than 
side term
  ∑[j=1,n]{1/(z-rj)} ---eqn.AU019
and eqn.14.54 great than side term
  P'(z)/P(z) ---eqn.AU020
How can they be equal?

<a name="ch06f051">　Index begin　Index this file
Here use a simple numerical example
to illustrate.
Assume 
  P(x) = (x-1)(x-2)(x-3) ---eqn.AU021
real number root IS complex number 
root. Then
  P'(x)= [d(x-1)/dx]*(x-2)(x-3)
        +(x-1)*[d(x-2)/dx]*(x-3)
        +(x-1)(x-2)*[d(x-3)/dx]
  P'(x)= (x-2)(x-3)+(x-1)*(x-3)
        +(x-1)(x-2) ---eqn.AU022

<a name="ch06f052">
  P'(z)/P(z)=
   [(x-2)(x-3)+(x-1)(x-3)+(x-1)(x-2)]
   /[(x-1)(x-2)(x-3)]

  P'(z)/P(z) ---eqn.AU023
  = 1/(x-1) + 1/(x-2) + 1/(x-3)
  = ∑[j=1,3]{1/(x-rj)}

<a name="ch06f053">
If the root is not simple 1,2,3. If
root is 1/2, 1/3, 1/4 then highest 
term coefficient an is not one.
But an still cancel out from
P'(z)/P(z). Because P'(z) and P(z)
both factor out same an.
cancel to one.
With eqn.AU023 illustration, it is
easy to understand eqn.AU019 and
eqn.AU020 are same thing.

<a name="ch06f054">
[[
for a z which is outside of the
convex hull H
]]
Make sure 1/(x-rj) or 1/(z-rj)
is not 1/0.

Exercise 6.13 solution is done.
2010-01-17-16-25 stop





<a name="ch06f055">　Index begin　Index this file
2010-01-17-18-17 start
■　Exercise 6.14 problem statement
　　textbook page 103
（A Polynomial Lower Bound）
//Polynomial roots, coefficients.

Given that the zeros of the polynomial
  P(z)=anzn+...+a1z+a0 ---eqn.AU024
are contained in the unit disk
  U={z:|z|≦1} ---eqn.AU025
<a name="ch06f056">
show that one has
  n|an|1/n|P(z)|(n-1)/n√[1-|z|-2]
  ≦|P'(z)| ---eqn.6.27
for all z not in U
2010-01-17-18-26 stop

<a name="ch06f058">
2010-01-17-18-59 start
■　Exercise 6.14 hint
　　textbook page 252
If 2ψ is the viewing angle determined 
by U when viewed from z not in U, 
then we have
  1 = |z|sin(ψ)  ---eqn.AU026
so Pythagoras's theorem tells us that
  cos(ψ)=[1-|z|-2]1/2 ---eqn.AU027
The target inequality (6.27) then
follows directly from Wilf's bound
(6.26)
2010-01-17-19-04 stop








<a name="ch06f059">　Index begin　Index this file
2010-01-17-19-07 start
■　Exercise 6.14 solution


Please click [Draw Ex.6.14] button
for unit disk drawing. All complex 
roots stay in side the unit disk.
Point z is not a root and z stay 
out side of unit disk. Drawing put
z at (-2,0i)

<a name="ch06f060">
From point z to coordinate original
point (0,0) is z. then
  |z| = |(-2,0i)| = 2 ---eqn.AU028
in this numerical example.

From drawing it is easy see that 
  |z|*sin(ψ)= disk radius = 1 ---eqn.AU026
<a name="ch06f061">
We have 
  sin(ψ) = 1/|z| ---eqn.AU029
Pythagoras's theorem tells us that
  cos(ψ)*cos(ψ) + sin(ψ)*sin(ψ) = 1 ---eqn.AU030
Put eqn.AU029 to eqn.AU030 solve for
cos(ψ) get
  cos(ψ)*cos(ψ) + 1/[|z|*|z|] = 1 
  cos(ψ)*cos(ψ) = 1 - 1/[|z|*|z|]
  cos(ψ)*cos(ψ) = [1-|z|-2]
take square root get
  cos(ψ)=[1-|z|-2]1/2 ---eqn.AU027

<a name="ch06f062">
Put cos(ψ) to eqn.6.26 get

<a name="ch06f063">
2010-01-17-19-30 here
eqn.6.27 right side has only |P'(z)|
then eqn.AU031 right side keep only
|P'(z)|. Move other terms to left.
When work with inequality and move
terms cross inequality, must check
the moved term is positive or negative.
Move positive term, inequality same. 
Move negative term, inequality reverse.
<a name="ch06f064">
For z out side of unit disk, z is not
a root of P(z), then |P(z)|＞0
For z out side of unit disk |z|＞1
 |z|*|z|＞1
 1/[|z|*|z|]＜1
 1-1/[|z|*|z|]＞0
 √(1-1/[|z|*|z|])＞0
Also polynomial order n >0. Everything
positive, we can move safely.
Write eqn.AU031 as

From eqn.AU032 to eqn.6.27 that is
your home work.
2010-01-17-19-50 stop



<a name="ch06f066">　Index begin　Index this file
2010-01-17-21-12 start
■　Exercise 6.15 problem statement
　　textbook page 103
（A Complex Mean Product Theorem）

Show that if 0＜r＜1 and if the complex
number z1,z2,...,zn are in the disk
<a name="ch06f067">
  D={z: |z|≦r} ---eqn.AU033
then there exists a z0 in D such that
  ∏[j=1,n](1+zj) = (1+z0)n ---eqn.6.28
2010-01-17-21-20 here


<a name="ch06f068">
2009-02-01-15-22 LiuHH access
http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/CSMC_Errata.pdf
it has
(Exercise 6.15, not Exercise 6.14)
[[
<a name="ch06f069">
Pages 103 and 252 regarding 
Exercise 6.14: Chapman also notes 
that the solution offered does not 
address the question as stated but 
rather its analogue with r = 1. The 
easy fix here is just to restrict 
the problem to the case of r = 1. 
Still, I suspect that the problem 
is correct as posed but one needs 
to do a scaling.
This is also on my "to fix" list.
]]
2010-01-17-21-24 record







<a name="ch06f070">　Index begin　Index this file
2010-01-17-21-27 start
■　Exercise 6.15 hint
　　textbook page 252
This is American Mathematical Monthly
Problem 10940 posed by V Nievergelt.
We consider the solution by A Nakhash.
<a name="ch06f071">
The disk 
  D0={z: |1-z|≦1} ---eqn.AU034
//Compare with
//D={z: |z|≦r} ---eqn.AU033
in polar coordinate is
  {reiθ: 0≦r≦2*cos(θ), -π/2＜θ＜π/2} ---eqn.AU035
<a name="ch06f072">
so for each j we can write 
  1+zj as rjeiθj ---eqn.AU036
where
  -π/2＜θ＜π/2  ---eqn.AU037
and where
  rj≦2*cos(θj) ---eqn.AU038
<a name="ch06f073">
It is immediately that
  z0=-1+(r1r2...rn)1/n ---eqn.AU039
       *exp(i(θ1+θ2+...+θn)/n)
solves Nievergelt equation (6.28),
and to prove that z0 in D it suffices 
to show 1+z0 in D0; equivalently
<a name="ch06f074">
we need to show
  (r1r2...rn)1/n≦2cos((θ1+θ2+...+θn)/n) ---eqn.AU040
Since (r1r2...rn)1/n is bounded by
((2*cos(θ1))(2*cos(θ2))...(2*cos(θn)))1/n
It therefore suffices to show that
  ((2*cos(θ1))(2*cos(θ2))...(2*cos(θn)))1/n
  ≦2cos((θ1+θ2+...+θn)/n) ---eqn.AU041
<a name="ch06f075">
and this follows the concavity of
  f(x)=log(cos(x))
  f(θ)=log(cos(θ)) ---eqn.AU042
on   -π/2＜θ＜π/2  ---eqn.AU037
together with Jensen's inequality.
2010-01-17-21-58 stop

<a name="ch06f076">
  f(x)  = ---eqn.AU043
log(cos(x))
  f'(x) = ---eqn.AU044
-sin(x)/cos(x)
  f''(x)= ---eqn.AU045
-1-sin(x)*sin(x)/cos(x)/cos(x)

Exercise 6.15 concave function drawing
please click here
Please click [17] button.
2010-01-17-22-10 here

<a name="ch06f078">　Index begin　Index this file
2010-01-18-16-16 start
■　Exercise 6.15 solution


Please pay attention to two disks
  D0={z: |1-z|≦1} ---eqn.AU034
and
  D={z: |z|≦r} ---eqn.AU033
<a name="ch06f079">
D0 center at (x,y)=(1,0), radius=1
D  center at (x,y)=(0,0), radius=1
Please click the button "Draw circle
at (1,0)"
Red circle is D, given in problem
statement
Blue circle is D0, given in hint

<a name="ch06f080">
Compare the polar coordinates:
Red:  radius=1, angle in [0,2*PI]
Blue: r=2*cos(θ), -π/2＜θ＜π/2

Point zj is in red circle D
Point rjeiθj is in blue circle D0.
<a name="ch06f081">
A point P who is in both D and D0
its phase angle must be in first
quadrant or fourth quadrant, that
is θ in -π/2＜θ＜π/2.
(How can a second/third quadrant
 point stay in blue circle?)
Please pay attention to that blue
circle is defined in -π/2＜θ＜π/2.
// 2010-01-21-18-22 gray color above text.
// Above say points in 1st/4th quadrant
// which is wrong. Points are in all
// four quadrants and in unit disk.

<a name="ch06f082">
P relative to D is expressed as zj
P relative to D0 is written as rjeiθj
D parallel shift on real axis one
unit become D0, therefore zj add one
become D0 member. Relative to D0,
we can write equivalence as next
  1+zj = rjeiθj ---eqn.AU036

<a name="ch06f083">　Index begin　Index this file
Exercise 6.15 hint say
[[
It is immediately that
  z0=-1+(r1r2...rn)1/n ---eqn.AU039
       *exp(i(θ1+θ2+...+θn)/n)
solves Nievergelt equation (6.28),
]]
rk and θk are all used by D0 member.
z0 is a D member. They are in two 
different circle,
Why "immediately" ?

<a name="ch06f084">
Let us re-write eqn.AU039 as next
  1+z0=+(r1r2...rn)1/n ---eqn.AU046
       *exp(i(θ1+θ2+...+θn)/n)
Now 1+z0 is a member of D0 (not D)
eqn.AU046 is an all D0 equation.
1+z0 is a Complex Mean Product point 
of all rjeiθj for j=1 to n.

<a name="ch06f085">
2010-01-18-17-03 here, wondering
WHY ?
If we can take Complex Mean Product 
point in D0,
why we can not take Complex Mean 
Product point in D ?
<a name="ch06f086">
Exercise 6.15 problem statement
did not exclude points in D and in 
second/third quadrant. What happen 
if some points are there?
If points are in red circle and in
second/third quadrant, those points
are NOT in blue circle !!
Wondering !
2010-01-18-17-08 
//2010-01-21-18-31 gray above text.
//all 2nd/3rd quadrant notes are grayed

<a name="ch06f087">
2010-01-18-17-13
Take n th power to eqn.AU046, we get
eqn.6.28 This is the reason hint say
[[
It is immediately that
]]

<a name="ch06f088">
LiuHH need time to think above 
"why, wondering" question.

Exercise 6.15 solution is NOT DONE.
2010-01-18-17-21 stop

<a name="ch06f089">
2010-01-21-18-50
When we work with real number
sequence and find their GM, we
always limit to non-negative real.
Because negative to 1/2 or 1/n power
are complex. In Exercise 6.15 we work
with complex number. Complex to 1/2
power is still complex. Why shift 
from unit disk D to real side D0 ?
They are complex any way!! wondering.
2010-01-21-18-54



<a name="ch06f090">　Index begin　Index this file
2010-01-18-18-20 start
■　Exercise 6.16 problem statement
　　textbook page 104
（Shapiro's Cyclic Sum Inequality）

<a name="ch06f091">
Show that for positive a1,a2,a3 and a4
One has the bound

<a name="ch06f092">
2010-01-18-18-28 here
Incidentally, the review of Bushell 
(1994) provides a great deal of infor-
mation about the inequalities of the
form

<a name="ch06f094">
2010-01-18-18-37 here
This bound is known to fail for n≧ 25,
yet the precise set of n for which it
is valid has not yet been determined.
2010-01-18-18-38 stop






<a name="ch06f095">　Index begin　Index this file
2010-01-18-18-40 start
■　Exercise 6.16 hint
　　textbook page 252

A nice solution using Jensen's inequality
for
  f(x)=1/x  ---eqn.AU048
was given by Robert Israel in the 
sci.math newsgroup in 1999. If we set
  S = a1＋a2＋a3＋a4 ---eqn.AU049
and let C denote the sum on the right
hand side of the bound (6.29), then
<a name="ch06f096">
Jensen's with
  pj=aj/S ---eqn.AU050
and
  x1=a2+a3 ---eqn.AU051
  x2=a3+a4 ---eqn.AU052
  x3=a4+a1 ---eqn.AU053
  x4=a1+a2 ---eqn.AU054
<a name="ch06f097">
gives us
  C/S ≧ {D/S}-1 ---eqn.AU055
or
  C ≧ S2/D ---eqn.AU056
where one has set //define D here
  D=a1(a2+a3)+a2(a3+a4)
   +a3(a4+a1)+a4(a1+a2) ---eqn.AU057
<a name="ch06f098">
Now it is easy to check that 
  S2-2D = (a1-a3)2+(a2-a4)2 ＞ 0 ---eqn.AU058
and this lucky fact suffice to
complete the solution.
2010-01-18-18-56 stop




<a name="ch06f099">　Index begin　Index this file
2010-01-18-19-17 start
■　Exercise 6.16 solution


Jensen's inequality is next

<a name="ch06f100">
Jensen's inequality says
for a convex function 
  AM in DOMAIN ≦ AM in RANGE ---eqn.AU059
Domain and range are different.

Exercise 6.16 hint suggest use the
function
  f(x)=1/x  ---eqn.AU048
as Jensen's inequality base function.
Whether f(x)=1/x is convex?
It is convex if and only if f''(x)＞0
2010-01-18-19-28 stop

<a name="ch06f102">
2010-01-18-19-53 start
Put eqn.AU059 and eqn.AU048 together
we see
  1/[(x1+x2+x3+x4)/4] ---eqn.AU060
  ≦ [1/x1 + 1/x2 + 1/x3 + 1/x4]/4
eqn.AU060 is
  AM in DOMAIN ≦ AM in RANGE ---eqn.AU059
We need modify eqn.AU060. Because
eqn.AU060 use equal weight 1/4, but
problem use unequal weight eqn.AU050.
Also problem use x1,x2,x3,x4 defined 
in eqn.AU051 to eqn.AU054.

<a name="ch06f103">
Let us see AM in DOMAIN first
change 1/[(x1+x2+x3+x4)/4]
to generalized weight (not 1/4)
 AM_in_DOMAIN=1/[(p1x1+p2x2+p3x3+p4x4)] ---eqn.AU061
Where pk sum to one is necessary.
Put eqn.AU049 and eqn.AU050 together
<a name="ch06f104">　Index begin　Index this file
find pj as
  p1=a1/(a1＋a2＋a3＋a4) ---eqn.AU062
  p2=a2/(a1＋a2＋a3＋a4) ---eqn.AU063
  p3=a3/(a1＋a2＋a3＋a4) ---eqn.AU064
  p4=a4/(a1＋a2＋a3＋a4) ---eqn.AU065
Main point is to say that
  p1+p2+p3+p4=1 ---eqn.AU066
generalized weight or equal weight
sum to one is a necessary condition.
Here confirmed that ∑pj=1

<a name="ch06f105">
Now put p's and x's into eqn.AU061
p's are defined in eqn.AU062 to AU065
x's are defined in eqn.AU051 to AU054
The denominator of AM_in_DOMAIN is
  (p1x1+p2x2+p3x3+p4x4) ---eqn.AU067
  =(a2+a3)*a1/(a1＋a2＋a3＋a4)
  +(a3+a4)*a2/(a1＋a2＋a3＋a4)
  +(a4+a1)*a3/(a1＋a2＋a3＋a4)
  +(a1+a2)*a4/(a1＋a2＋a3＋a4)
<a name="ch06f106">
eqn.AU067 numerator is defined as D
see eqn.AU057.
eqn.AU067 denominator is defined as S
see eqn.AU049. Then
  eqn.AU067 = D/S ---eqn.AU068

<a name="ch06f107">
This whole thing eqn.AU067 is denominator
of AM_in_DOMAIN eqn.AU061. Put D/S to
denominator, we see 
AM_in_DOMAIN eqn.AU061=1/(D/S)=(D/S)-1
This is eqn.AU055 less than side.

Above is AM_in_DOMAIN
Below is AM_in_RANGE.
2010-01-18-20-27 here

<a name="ch06f108">
Please see eqn.AU060 greater than 
side (next line)
[1/x1 + 1/x2 + 1/x3 + 1/x4]/4 ---eqn.AU069
This is equal weight AM_in_RANGE.
We have unequal weight AM_in_RANGE
as following
  AM_in_RANGE= ---eqn.AU070
  p1/x1 + p2/x2 + p3/x3 + p4/x4
here pj is generalized weight
(pj sum to one is confirmed)
1/xj is function 1/x evaluation
at x=xj.

<a name="ch06f109">　Index begin　Index this file
Again put p's and x's into eqn.AU070
find
  AM_in_RANGE= ---eqn.AU071
   p1/x1
 + p2/x2
 + p3/x3
 + p4/x4
  =
   a1/[S*(a2+a3)]
 + a2/[S*(a3+a4)]
 + a3/[S*(a4+a1)]
 + a4/[S*(a1+a2)]

<a name="ch06f110">
  AM_in_RANGE= ---eqn.AU072
 { a1/[(a2+a3)]
 + a2/[(a3+a4)]
 + a3/[(a4+a1)]
 + a4/[(a1+a2)]
 }/S
Four terms in eqn.AU072 {.....} is 
defined to be C. We have
  AM_in_RANGE=C/S ---eqn.AU073

<a name="ch06f111">
Jensen's inequality says
for a convex function 
  AM in DOMAIN ≦ AM in RANGE ---eqn.AU059
we get 
  1/(D/S) ≦ C/S
re-write as
  S*S/D ≦ C ---eqn.AU074
eqn.AU074 and eqn.AU056 are same thing.
2010-01-18-20-58 here

<a name="ch06f112">
Jensen's inequality give us S*S/D ≦ C

Target equation eqn.6.29 ask to prove
  2≦C ---eqn.AU075
In eqn.AU075, C is positive term.
Expect minimum of C is 2. 

In the link
  2 ≦ S*S/D ≦ C ---eqn.AU076
S*S/D ≦ C is Jensen's inequality.
 2 ≦ S*S/D is next to prove.

<a name="ch06f113">
if 2 ≦ S*S/D is true, then
   2 ≦ C is true for sure.

Now verify  2 ?≦? S*S/D 
that is
  0 ?≦? S*S - 2*D  ---eqn.AU077

<a name="ch06f114">　Index begin　Index this file
  S*S - 2*D = ---eqn.AU078
  (a1＋a2＋a3＋a4)2 //S defined, D defined
  -2*a1(a2+a3)-2*a2(a3+a4)
  -2*a3(a4+a1)-2*a4(a1+a2)
  =
  +[a1a1+a2a2+a3a3+a4a4]
  +2*[a1a2+a1a3+a1a4+a2a3+a2a4+a3a4]
  -2*[a1a2+a1a3+a2a3+a2a4
    +a3a4+a3a1+a4a1+a4a2]
<a name="ch06f115">
  =
  +[a1a1+a2a2+a3a3+a4a4]
  +2*[0]
  -2*[0+a3a1+a4a2]
  =
  +[a3-a1]2
  +[a4-a2]2

  S*S - 2*D ≧ 0 ---eqn.AU079
2010-01-18-21-26 here

<a name="ch06f116">
  We showed that 
  S*S/D ≧ 2 ---eqn.AU080
Jensen's inequality give us S*S/D ≦ C
All together is
  C ≧ S*S/D ≧ 2 ---eqn.AU081
Exercise 6.16 problem solved.
2010-01-18-21-47 stop



<a name="ch06f117">　Index begin　Index this file
2010-01-19-09-27 start
■　Exercise 6.17 problem statement
　　textbook page 104
（The Three Chord Lemma）

Show that if f:[a,b]→Real is convex,
and a＜x＜b then one has

<a name="ch06f119">
2010-01-19-09-35 here
As the next two exercises suggest, 
this bound is the key to some of the
most basic regularity properties of
convex functions.
2010-01-19-09-36 stop

Please goto Ex.0618 Drawing for a 
drawing relate to Exercise 6.17.
Figure 6.1 frame (A) or frame (B) 
both explain Exercise 6.17.







<a name="ch06f120">
2010-01-19-09-39 start
■　Exercise 6.17 hint
　　textbook page 253

By interpolation and convexity one has

<a name="ch06f122">
This gives us the left inequality 
of (6.30) and the right is proved
in the same way.
2010-01-19-09-59 stop

Textbook use
[[
This gives us the second inequality 
of (6.30) and the second is proved
in the same way.
]]
LiuHH change "the second" to
"the left" and "the right"
to make difference.
2010-01-19-10-16 here




2010-01-19-10-18 start
<a name="ch06f123">　Index begin　Index this file
■　Exercise 6.17 solution


Problem statement given
f:[a,b]→Real is convex, and a＜x＜b 
Convex definition is
AM in DOMAIN ≦ AM in RANGE
<a name="ch06f124">
In eqn.AU082 left half
  x = a*(b-x)/(b-a) + b*(x-a)/(b-a) ---eqn.AU083
this is AM in DOMAIN.
AM is Arithmetic Mean. 
Points a and b are in domain.
Point a has a weight (b-x)/(b-a)
Point b has a weight (x-a)/(b-a)
<a name="ch06f125">
These two weights sum to one
  (b-x)/(b-a) + (x-a)/(b-a) = 1 ---eqn.AU084
(Please pay attention to that a＜x＜b )
Weights sum to one is a necessary 
condition. This condition is true.
<a name="ch06f126">
We can not compare x with AM in RANGE
directly. Next equation is WRONG
  x_AM
   =
  a*(b-x)/(b-a) + b*(x-a)/(b-a) 
   ≦  ---eqn.AU085 WRONG
  f(a)*(b-x)/(b-a) + f(b)*(x-a)/(b-a) 
Refer to distance, velocity, acceleration.
"eqn.AU085 WRONG" is time ≦ distance.
which is not acceptable.
<a name="ch06f127">
But we can compare f(x_AM) with
AM_RANGE. It is 
distance compare with distance
this comparison is reasonable.
Correct comparison is eqn.AU082 right
half. Copy as next
<a name="ch06f128">
  f(x_AM)
   =
  f(a*(b-x)/(b-a) + b*(x-a)/(b-a) )
   ≦  ---eqn.AU086 CORRECT
  f(a)*(b-x)/(b-a) + f(b)*(x-a)/(b-a) 
<a name="ch06f129">
We get "eqn.AU086 CORRECT" from 
convexity assumption. The following
is easy. In "eqn.AU086 CORRECT" both
side subtracting f(a), and a small 
re-arrangement 
// you need do some simple work. Less
// than side is simply f(x)-f(a) This
// 'x' is defined in eqn.AU083
get eqn.14.56. which is the left 
inequality of (6.30).

(6.30) right inequality is same as above.
2010-01-19-11-03 stop





<a name="ch06f130">　Index begin　Index this file
2010-01-19-15-19 start
■　Exercise 6.18 problem statement
　　textbook page 104
（Near Differentiability of Convex Function）

Use the Three Chord Lemma to show
that for convex f:[a,b]→Real and
a＜x＜b one has the existence of 
the finite limits

2010-01-19-15-40 here

<a name="ch06f132a">
2010-01-22-12-09 proofread note start
eqn.AU087 is [f(x+h)－f(x)]/[(x+h)－(x)]
similarly
eqn.AU088 is [f(x－h)－f(x)]/[(x－h)－(x)]
then eqn.AU088
should be  [f(x－h)－f(x)]/[－h]
instead of [f(x－h)－f(x)]/[＋h]
2010-01-22-12-13 proofread note stop








<a name="ch06f133">
2010-01-19-15-41 start
■　Exercise 6.18 hint
　　textbook page 253

Let
  g(h)={f(x+h) - f(x)}/h ---eqn.AU089
and check from the Three Chord Lemma 
that for
  0＜h1＜h2 ---eqn.AU090
one has
  g(h1) ≦ g(h2) ---eqn.AU091
<a name="ch06f134">
Next choose y with
  a＜y＜x ---eqn.AU092
and use the Three Chord Lemma to
check that 
  －∞＜{f(x)-f(y)}/(x-y)≦g(h) ---eqn.AU093
for all h＞0. The monotonicity and 
boundedness g(h) guarantee that g(h) 
has finite limit as h→0. This gives
us the first half of the problem,
and the second half almost identical.
2010-01-19-15-51 stop

<a name="ch06f135">
2010-01-19-18-26 start
■　Exercise 6.18 solution


Three Chord Lemma is
[[
if f:[a,b]→Real is convex, and a＜x＜b
then //next line ---eqn.AU094
 　 chord(x,a) slope //left interval
 ≦ chord(b,a) slope //whole interval
 ≦ chord(b,x) slope //right interval
]]
eqn.AU094 is simplified eqn.6.30
Please click [Draw Ex.0618] and see
frame (B)
<a name="ch06f136">
There are three chords in frame (B).
Each chord has a slope. 
Exercise 6.18 work with chord slope.
Frame (B) has y axis and x axis.
y axis is function value.
x axis is independent variable value.
  Chord slope = ---eqn.AU095
  y_value difference/x_value difference

<a name="ch06f137">
Hint define g(h) as next
  g(h)={f(x+h) - f(x)}/h ---eqn.AU089
y_value difference is f(x+h) - f(x)
x_value difference is (x+h)-(x)=h
Two differences ratio is slope.
Therefore, g(h) is a slope equation.

<a name="ch06f138">
Hint continue say
[[
  0＜h1＜h2 ---eqn.AU090
]]
Please goto E0618Draw click [Draw
 Ex.0618] and see frame (B)
'0' in 0＜h1＜h2 is point 'a' in (B)
'h1' in 0＜h1＜h2 is x-step 'ax' in (B)
'h2' in 0＜h1＜h2 is x-step 'ab' in (B)
Above 'point' and 'x-step' are all on
x-axis, not involve y-axis.



<a name="ch06f139">　Index begin　Index this file
Hint continue say
[[
one has
  g(h1) ≦ g(h2) ---eqn.AU091
]]
g(h1) is chord LM slope
g(h2) is chord LN slope
Slope involve y-axis. (see eqn.AU095)

<a name="ch06f140">
In frame (B) domain order is a＜x＜b.
For a＜x＜b, three Chord Lemma say :
  chord(a,x) slope ≦ chord(a,b) slope
this is
  g(h1) ≦ g(h2)  ---eqn.AU091
[[
<a name="ch06f140A">
2010-01-21-22-38 add start
From mean value theorem: if curve
is continuous and differentiable.
There is a point s in [a,x] such 
that tangent at s has slope same 
as chord(a,x) slope.
There is a point t in [a,b] such 
that tangent at t has slope same 
as chord(a,b) slope.
<a name="ch06f140B">
Since a＜x＜b, so s＜t. From convex
property (not from three Chord Lemma)
tangent s slope ＜ tangent t slope
same result as eqn.AU091
2010-01-21-22-44 add stop
]]
<a name="ch06f141">
eqn.AU091 main point is that 
For a convex function f(x) if a＜x＜b 
then when x increase from x=a to x=b
the slope has monotonicity 
(curve tangent slope monotone increase)

<a name="ch06f142">
Please goto tute0022.htm
Click convex example buttons
0, 1, 4, 5, 7, 9, 12, 14, 15
All convex curve slope function
(black curve) are monotone increase

<a name="ch06f143">
"black curves monotone increase" is
a graphic illustration for
  g(h1) ≦ g(h2)  ---eqn.AU091
(certainly, for concave function 
 black curves monotone decrease)

Above: slope has monotonicity 
<a name="ch06f144">　Index begin　Index this file
Below: slope is bounded.

Hint continue say
[[
Next choose y with
  a＜y＜x ---eqn.AU092
and use the Three Chord Lemma to
check that 
  －∞＜{f(x)-f(y)}/(x-y)≦g(h) ---eqn.AU093
for all h＞0.
]]

<a name="ch06f145">
For domain relation l＜m＜n
we know
   lm slope   ≦ ln slope ≦ mn slope
left interval   whole int.  right interval

in the case
  a＜y＜x ---eqn.AU092
we expect yx slope is the greatest
Because   this  yx in a＜y＜x
compatible with mn in l＜m＜n
mn slope is greatest, must have
yx slope is the greatest.

<a name="ch06f146">
In eqn.AU093, why the greatest yx slope
is defeated by g(h) ? see
  －∞＜{f(x)-f(y)}/(x-y)≦g(h) ---eqn.AU093
2010-01-19-19-37 here

<a name="ch06f147">
Better explanation is that the domain 
section is not just 
  a＜y＜x ---eqn.AU092
but it is wider
  a＜y＜x＜b ---eqn.AU096
g(h) is xb slope, which is real greatest.
Then what is
  a＜y＜x ---eqn.AU092
for?
<a name="ch06f148">
Here a＜y, not allow y touch a.
y must be greater than a. 
This requirement a＜y make sure yx slope
not reach -infinity.

<a name="ch06f149">
Please goto E0618Draw click [Draw
 Ex.0618] and see frame (C) where
f(x) = 1/cos(x) and
 x in (-PI/2, +PI/2)
a＜y＜x  not allow point go near -PI/2

<a name="ch06f150">
With monotonicity and boundedness
in hand, hint say
[[
The monotonicity and 
boundedness g(h) guarantee that g(h) 
has finite limit as h→0.
]]
and solve problem.

LiuHH is learning. 
Exercise 6.18 solution heavily rely
upon Exercise 6.18 hint.
2010-01-19-19-51 stop





<a name="ch06f151">　Index begin　Index this file
2010-01-19-21-22 start
■　Exercise 6.19 problem statement
　　textbook page 104
（Ratio Bounds and Linear Minorants）

For convex f:[a,b]→Real is convex, 
and a＜x＜y＜b, show that one has

<a name="ch06f153">
2010-01-19-21-35 here
In particular, note that for each 
θ∈[f'－(x),f'＋(x)], one has the 
bound
  f(y)≧f(x)+(y-x)θ ---eqn.6.32
  for all y∈[a,b]
<a name="ch06f154">
The linear lower bound (6.32) is more
effective that its simplicity would
suggest, and it has some notable
consequences. In the next chapter
we will find that it yields and
exceptionally efficient proof of
Jensen's inequality.
2010-01-19-21-40 here






<a name="ch06f155">　Index begin　Index this file
2010-01-19-21-42 start
■　Exercise 6.19 hint
　　textbook page 253

This is just more handy work of the
Three Chord Lemma which gives us
for 0＜s and 0＜t with y-s∈I and
y+t∈I that 
  {f(y)-f(y-s)}/s ≦
  {f(y+t)-f(y)}/t ---eqn.AU097
<a name="ch06f156">
From Exercise 6.18 we have that 
finite limits as s,t→0, and these
limits are f＇－(y) and f＇＋(y) 
respectively.
This gives us
  f＇－(y) ≦ f＇＋(y) ---eqn.AU098
and the other bounds are no harder. 
Incidentally, the bound f＇－(y) ≦ f＇＋(y)
may be regarded as an "infinitesimal" 
version of the Three Chord Lemma

<a name="ch06f157">
For
  a＜x≦s≦t≦y＜b  ---eqn.AU099
and
  M=max{|f'+(x)|, |f'-(y)|} ---eqn.AU100
the bound (6.31) gives us 
  |f(t)-f(s)|≦M|t-s| ---eqn.AU101
which is more than we need to say
that f is continuous.
2010-01-19-21-58 stop



2010-01-20-12-50 start
<a name="ch06f158">
■　Exercise 6.19 discussion

Exercise 6.19 work with tangent
slope of a convex function.
Simply speak, it is
For a convex function, if variable
move from domain left end to right
end. Function tangent slope always
increase or no change. Slope never
reduce.
<a name="ch06f159">
Exercise 6.19 main point is to study
composite convex function with two
corners, where slope discontinuous.
2010-01-20-12-54 here

<a name="ch06f160">
2010-01-20-17-37 start
f(y) is function f(x) evaluation 
at x=y

f＇－(y) is function f(x) derivative
df(x)/dx evaluation at x=y and at
x=y left end (variable x<y constant)

f＇＋(y) is function f(x) derivative
df(x)/dx evaluation at x=y and at
x=y right end.

<a name="ch06f161">
If f(x) is a single function
(NOT composited by two or more
 function) and if f(x) is
differentiable at all points.
Then every point left derivative
and right derivative are same.

<a name="ch06f162">
Now eqn.6.31 emphasize f＇－(y) and 
f＇＋(y) . This exercise 6.19 
target at composite function with
corner.

<a name="ch06f163">
What is the meaning of eqn.6.31 ?
Please click E0619Draw [AA Draw]
You will find a composite function
which is convex and has two corners.

<a name="ch06f164">
blue curve:  f(x)=x*x
red curve:   f(x)=x*log(x)
black curve: f(x)=exp(x)
black curve and black tangent cannot 
be distinguished.
'x' is a joint point (not a variable
here) between blue and red.
'y' is a joint point between red and
black.

<a name="ch06f165">　Index begin　Index this file
Blue line is a tangent to blue curve
at point 'x'
Red solid line is a tangent to red 
curve at point 'x'
'x' left derivative is Blue line slope.
'x' right derivative is red line slope.

<a name="ch06f166">
Red dash line is a tangent to red 
curve at point 'y'
Black line and black curve coincide
in this drawing.
Black line is a tangent to black curve 
at point 'y'

<a name="ch06f167">
'y' left derivative is red dash line
    slope.
'y' right derivative is black line
    slope.

Red dot line connect point 'x' and 'y'.

A point move on x axis from left to
right (function variable value increase)
The related tangent slope increase.

<a name="ch06f168">
eqn.6.31 say 
blue line slope ≦ red solid line slope
and
red solid line slope ≦ red dot line slope
and
red dot line slope ≦ red dash line slope
and
red dash line slope ≦ black line slope
Above description has four '≦', 
they match four '≦' in eqn.6.31
they match four tangents in [AA Draw] too.

<a name="ch06f169">
This tangent slope monotone increase is
necessary for a convex function.

One more time
Please goto tute0022.htm
Click convex example buttons
0, 1, 4, 5, 7, 9, 12, 14, 15
All convex curve slope function
(black curve) are monotone increase.
This observation is consistent with
eqn.6.31

<a name="ch06f170">　Index begin　Index this file
What happen if violate eqn.6.31?
We need a counter example.

Violation means that function
variable increase and tangent slope
decrease. This will slow down curve
increase rate. Convexity is void.

<a name="ch06f171">
Please click E0619Draw [BB Draw]
(not AA button now, click BB) 
You will find a composite function
which has one corner and slope
decrease. Then red dotted chord 
can not always above function curve.

"chord not always above function curve"
dis-qualify function as convex.
2010-01-20-18-26 stop





<a name="ch06f172">
2010-01-20-18-27 start
■　Exercise 6.19 solution


We ahve Three Chord Lemma in hand.
Please goto E0618Draw click
[Draw Ex.0618] button. In frame (B),
take 'x' point as target point.

<a name="ch06f173">
When point 'a' approach 'x' (from left),
chord LM slope approach to 'x' point left 
derivative.
When point 'b' approach 'x' (from right),
chord MN slope approach to 'x' point right
derivative.

<a name="ch06f174">　Index begin　Index this file
Three Chord Lemma help us solve corner
left/right derivatives perfect.
If function is convex, Three Chord 
Lemma tell us the result
corner left derivative
 <=
corner right derivative
<a name="ch06f175">
eqn.AU097 is just a mathematical 
expression for the problem in hand.
's' in eqn.AU097 is 'ax' in (B) frame.
't' in eqn.AU097 is 'xb' in (B) frame.
We already proved Three Chord Lemma,
just use it to draw conclusion.

2010-01-20-18-40 here

<a name="ch06f176">
2010-01-20-18-52 start
About continuity, if f(x) is discontinue
at x=m, then f(x) convexity disappear.
For example f(x)=x*x is convex. Now cut
f(x)=x*x at x=0 and lift right half one
unit higher. That is
for x<0, g(x)=x*x
for x>0, g(x)=x*x+1
<a name="ch06f177">
This g(x) is discontinuous at x=0.
This g(x) is not convex. Because the
chord connect x=-0.5, g(-0.5)=0.25
and the point x=+0.5, g(+0.5)=1.25
below function curve for x in (0, 0.5).

<a name="ch06f178">
Function continuity is guaranteed by
the Three Chord Lemma.
Since Lemma exclude infinity slope.
2010-01-20-19-01 stop

<a name="ch06f179">　Index begin　Index this file
2010-01-21-11-29 start
The meaning of E0619Draw [AA Draw]
is the following

For a composite convex function with 
two corners, [AA Draw] say

<a name="ch06f180">
left corner left  derivative (tangent) slope
is less than or equal to
left corner right derivative slope
is less than or equal to
left corner to right corner chord slope
is less than or equal to
right corner left  derivative slope
is less than or equal to
right corner right derivative slope

<a name="ch06f181">
In mathematics symbol, it is eqn.6.31
repeat as following
    f'-(x) <= f'+(x)
 <= [f(y)－f(x)]/[y-x]
 <= f'-(y) <= f'+(y)
Attention: there are FOUR inequality sign.

<a name="ch06f182">
In color line, (please see [AA Draw])
it is
    solid Blue slope
 <= solid red slope
 <= dot red slope
 <= dash red slope
 <= black slope

2010-01-21-11-35 here
2010-01-21-12-24 start

<a name="ch06f183">　Index begin　Index this file
In number, slope from left to right is
 -2 ≦ -0.609437 ≦ 0.727692
    ≦ 1.405465 ≦ 7.3890
Attention: there are FOUR inequality sign.
Attention: slope number monotone increase.
Above numbers are NOT function values.
Above numbers are NOT variable values.
They are function value difference
divided by variable value difference.

<a name="ch06f184">
Above numbers are found as following
First slope
curve end at x=-1
f(x)=x*x, f'(x)=2*x
blue solid line slope = 2*x=2*(-1)=-2

Why blue solid  end  at x=-1 ,
but  red solid start at x=0.2 ? //9901211249

<a name="ch06f185">
Second slope
curve start at x=0.2
f(x)=x*log(x), f'(x)=log(x)+1
red solid line slope= log(0.2)+1= -0.609437

<a name="ch06f186">
Third slope
left  corner coord: 2.2, 1.678
right corner coord: 3.5, 2.624
slope=(2.624-1.678)/(3.5-2.2)=0.727692

<a name="ch06f187">
Fourth slope
curve end0= 1.5
f(x)=x*log(x), f'(x)=log(x)+1
red dash slope= log(1.5)+1=1.4054651

<a name="ch06f188">
Fifth slope
curve begin at x=2.0
f(x)=exp(x) , f'(x)=exp(x)
f'(2.0)=exp(2.0)=7.3890
black tangent line slope = 7.3890
black line and black curve exp(x)
coincide in this graph. 9901211220

2010-01-21-12-31 stop

<a name="ch06f189">　Index begin　Index this file
2010-01-21-12-50 start
Why blue solid  end  at x=-1 ,
but  red solid start at x=0.2 ?

Why red  dash   end  at x=1.5 ,
but black line begin at x=2.0 
yet they are continuous?

<a name="ch06f190">
This composite curve merged with three
curves
blue curve : f(x)=x*x
red curve  : f(x)=x*log(x)
black curve: f(x)=exp(x)
To make composite curve continuous,
each part has a parallel translation
constants. Although each has different
begin x and end x, but translation
constants make the final curve a
continuous curve.

<a name="ch06f191">
LiuHH headache for few hours to create
[AA Draw] and [BB Draw], hope it will
help you understand the meaning of
eqn.6.31.
2010-01-21-12-58 stop

<a name="ch06f192">
2010-01-21-16-44 start
How to get complex polynomial 
coefficients from given complex
roots?

Assume polynomial roots are following
FIVE complex numbers
1+i
1-2i
2+i
2+i
3-2i
<a name="ch06f193">
To find complex polynomial coefficients
please go to polyroot.htm at
http://freeman2.com/polyroot.htm

Step 1
Fill given five roots to [Box 1, answer]
(although it is not answer) 

Step 2
click [given root find coef] button.

<a name="ch06f194">
Now Box 1 has coefficients
-57-1i  //constant
99-18i  //z^1 coef.
-78+16i //z^2 coef.
36-6i   //z^3 coef.
-9+1i   //z^4 coef.
1+0i    //z^5 coef.

Given roots move to Box 2.
2010-01-21-17-02 here

<a name="ch06f195">
To verify whether this coef. set has
roots we have in hand. Do the next

Step 1
in the line
"max.poly.order [  ] 2 to 100 "
fill 5 (we have five roots)
Click [05] is the same.

<a name="ch06f196">
Step 2
Fill given six coefficients to 
[Box 1, answer]
(although it is not answer) 
If you do verify, coeff. is already
in Box 1. 
If you have a new polynomial
You need to fill polynomial coef.
to box 1. Constant on top. Highest
power coef. at bottom. Zero coef.
must fill a '0'

<a name="ch06f197">
Step 3
click [Box 1 is coef fill to small box]
Program read from Box 1 and fill to
small boxes C00 to C05.
If you do hard work, fill six numbers
to six small boxes, same effect.

<a name="ch06f198">
Step 4
click [fast get root verify answer] button.
Answer show up at Box 1
They are roots.
Box 2 has verification. All roots 
evaluation with given polynomial
should get zero value. +/-1.e-7
is acceptable in most case.
2010-01-21-17-13 stop

2010-01-21-20-35 done proofread first time
2010-01-22-12-48 done proofread second time
first proofread found that in ch06f138 has 
error. Correct and re-do second proofread.
2010-01-22-16-57 done spelling check



========= Chapter six end here =========




<a name="docB001">　Index begin　Index this file
2010-01-20-19-21 start
■　About symbolCandidate
When copy B. O. Peirce integration table 
page 75, item 593
need to use 土 干 found "± ∓". In notepad
"∓" is "-+", but in browser "∓" is a 
square. Go find Chinese characters,
土 is 'earth', 'mud', read as 'tu',
 it can be used as "+-"
干 is 'interfere', 'concern', read as
  'gan' it can be used as "-+".
Added few more simple Chinese for 
possible symbol replacement.

<a name="docB002">
Jensen's Inequality concern convex and
concave. Related Chinese characters are
convex:凸　, concave:凹 both are picture
character.

<a name="docB003">
Another interesting picture character
is 山 if you do not see the picture,
here is its original form five thousand
years ago: ∆∆∆

<a name="docB004">
出 is 山 after 山, must be traveling.
First thing to do when travel is exit
from home. Now 出 is exit, not travel.
2010-01-20-19-46 stop



<a name="Copyright">　Index begin　Index this file
2009-06-19-10-48
If you are interested in inequality,
suggest you buy the book
The Cauchy-Schwarz Master Class
written by Prof. J. Michael Steele
The Cauchy-Schwarz Master Class is
this web page's textbook.

To buy textbook, that is to show thanks
to Prof. J.M. Steele's great work.
and it is also respect copyright law.
Thank you. Freeman 2009-06-19

The Cauchy-Schwarz Master Class
J. Michael Steele ★★★★★
ISBN 978-0-521-54677-5
2009-06-19-10-56